Question

A large operator of timeshare complexes requires anyone interested in making a purchase to first visit the site of interest. Historical data indicates that 20% of all potential purchasers select a day visit, 50% choose a one-night visit, and 30% opt for a two-night visit. In addition, 10% of day visitors ultimately make a purchase, 30% of one-night visitors buy a unit, and 20% of those visiting for two nights decide to buy. Suppose a visitor is randomly selected and is found to have made a purchase. How likely is it that this person made a day visit? A one-night visit? A two-night visit?

Answer 1

Answer:

135%is an absolute and persent

Answer 2

Given:

$20%$% of customers select a day visit

$50$% customers select one-night visit

$30$% customers for a two-night visit

$10%$% day visitors make a purchase

$30$% one night visitors buy a unit

$20$% visit for two nights decide to buy

To Find:

The probability that how likely is it that this person made a day visit

Step-by-step explanation:

Assume $A_{1}=$for the selection a day visit

$A_{2}=$for the selection of night visit

$A_{3}=$select a two-night visit

$B=$makes a purchase

The probabilities are

$P(A_{1})=0.20\\P(A_{2})=0.50\\P(A_{3})=0.30\\P(B/A_{1} )=0.10\\P(B/A_{2})=0.30\\P(B/A_{3})=0.20$

The probability that the visitor made the purchase is

$P(B)=P(B/A_{1})P(A_{1})+P(B/A_{2})P(A_{2})+P(B/A_{3})P(A_{3})\\=(0.20)\times(0.10)+(0.50)\times(0.30)+(0.30)\times(0.20)\\=0.02+0.15+0.6\\=0.23$

Use Bayes theorem

Therefore, the probability to made a day visit they made a purchase is

$P(A_{1}/B)=\dfrac{P(B/A_{1})\times P(A_{1}) }{P(B)}$

$=\dfrac{0.20\times0.10}{0.23}\\=\dfrac{0.20\times0.10}{0.23}\\\\=\dfrac{0.02}{0.23}$

$=0.087.$

The probability to made a day visit is 0.23.